$\mathbb{Z}/n\mathbb{Z}$ projective as $\mathbb{Z}/n\mathbb{Z}$-module

Question

$\mathbb{Z}/n\mathbb{Z}$ as $\mathbb{Z}$-module is not projective because isn't torsionfree, but is projective as $\mathbb{Z}/n\mathbb{Z}$ module ?

Answer 1

For every unital ring $R$, the free $R$-module generated by $n$ elements is $R^n$.

In particular, for $n=1$, we get that $R$ is free (generated by one element), hence it is also projective, as each $R$-morphisms $f:R\to M$ is uniquely determined by the arbitrary element $f(1)\in M$.

Answer 2

$\Bbb Z / n \Bbb Z$ is free as a $\Bbb Z / n \Bbb Z$-module, so it's projective.

Answer 3

Proofs of these kind of statements depend on the definitions being used.

Let $R$ be a ring.

Definition. An $R$-module $P$ is a projective $R$-module if for every surjective $R$-module homorphism $M\xrightarrow{f}P$ there exists an $R$-module homomorphism $P\xrightarrow{s}N$ such that $f\circ s=\DeclareMathOperator{id}{id}\id_P$.

The answer to your question then follows from the following proposition.

Proposition. $R$ is a projective $R$-module.

Proof. Let $M\xrightarrow{f}P$ be a surjective $R$-module homomorphism and let $m\in f^{-1}(\{1_R\})$. Let $R\xrightarrow{s}N$ be $s(r)=r\cdot m$ (check that this is an $R$-module homomorphism). Then $f(s(r))=f(r\cdot m)=r\cdot f(m)=r\cdot 1_R=r$. Hence $f\circ s=\id_R$ so that $R$ is a projective $R$-module as advertised. $\Box$